∫ cos 3 x d x = sin x − 1 3 sin 3 x + C d d x [ sin x − 1 3 sin 3 x + c ] d d x cos x − 1 3 ⋅ 3 sin x 2 cos x + 0 {\displaystyle {\begin{aligned}&\int \cos ^{3}xdx=\sin {x}-{\frac {1}{3}}\sin ^{3}x+C\\[2ex]&{\frac {d}{dx}}{[\sin {x}-{\frac {1}{3}}\sin ^{3}{x}+c]}&{\frac {d}{dx}}{\cos {x}-{\frac {1}{3}}\cdot 3\sin {x^{2}}\cos {x}+0}\end{aligned}}}
![{\displaystyle {\begin{aligned}&\int \cos ^{3}xdx=\sin {x}-{\frac {1}{3}}\sin ^{3}x+C\\[2ex]&{\frac {d}{dx}}{[\sin {x}-{\frac {1}{3}}\sin ^{3}{x}+c]}&{\frac {d}{dx}}{\cos {x}-{\frac {1}{3}}\cdot 3\sin {x^{2}}\cos {x}+0}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/f7a228dd37c0db6b3bcebaf569109b5d374076c3)